Hard
Maximum Sum of 3 Non-Overlapping Subarrays — Python
Full explanation · Time O(n) · Space O(n)
# Time: O(n)
# Space: O(n)
class Solution(object):
def maxSumOfThreeSubarrays(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: List[int]
"""
n = len(nums)
accu = [0]
for num in nums:
accu.append(accu[-1]+num)
left_pos = [0] * n
total = accu[k]-accu[0]
for i in xrange(k, n):
if accu[i+1]-accu[i+1-k] > total:
left_pos[i] = i+1-k
total = accu[i+1]-accu[i+1-k]
else:
left_pos[i] = left_pos[i-1]
right_pos = [n-k] * n
total = accu[n]-accu[n-k]
for i in reversed(xrange(n-k)):
if accu[i+k]-accu[i] > total:
right_pos[i] = i
total = accu[i+k]-accu[i]
else:
right_pos[i] = right_pos[i+1]
result, max_sum = [], 0
for i in xrange(k, n-2*k+1):
left, right = left_pos[i-1], right_pos[i+k]
total = (accu[i+k]-accu[i]) + \
(accu[left+k]-accu[left]) + \
(accu[right+k]-accu[right])
if total > max_sum:
max_sum = total
result = [left, i, right]
return result